Question

If $\int \frac{(1+e^x).e^x}{(1+e^x)^2+1}dx = ln(\sqrt{g(x)})+C$, and $g(1)=(1+e)^2+1$ then the value of $g(0)$ is equal to

Answer 1

5

Answer 2

To solve the given problem, we need to evaluate the integral and then use the given conditions to find the value of $g(0)$.

Step 1: Evaluate the integral The given integral is $\int \frac{(1+e^x).e^x}{(1+e^x)^2+1}dx$. Let's use the substitution method. Let $t = 1+e^x$. Differentiating both sides with respect to $x$, we get: $dt = e^x dx$.

Now, substitute $t$ and $dt$ into the integral: $\int \frac{t}{t^2+1} dt$ This integral is of the form $\int \frac{f'(x)}{f(x)} dx = \ln|f(x)| + C$, if we consider $f(t) = t^2+1$. The derivative of $t^2+1$ is $2t$. So, we need to adjust the numerator by a factor of 2. $\int \frac{t}{t^2+1} dt = \frac{1}{2} \int \frac{2t}{t^2+1} dt$ Now, apply the integration formula: $\frac{1}{2} \ln|t^2+1| + C$ Substitute back $t = 1+e^x$: $\frac{1}{2} \ln((1+e^x)^2+1) + C$ Since $(1+e^x)^2+1$ is always positive for all real $x$, we can remove the absolute value signs: $\frac{1}{2} \ln((1+e^x)^2+1) + C$

Step 2: Match the result with the given form We are given that the integral is equal to $ln(\sqrt{g(x)})+C$. Let's rewrite our result using logarithm properties ($a \ln b = \ln b^a$): $\frac{1}{2} \ln((1+e^x)^2+1) = \ln(((1+e^x)^2+1)^{1/2}) = \ln(\sqrt{(1+e^x)^2+1})$ Comparing this with $ln(\sqrt{g(x)})+C$, we can identify $g(x)$: $g(x) = (1+e^x)^2+1$

Step 3: Verify the given condition The problem states that $g(1)=(1+e)^2+1$. Let's check our derived $g(x)$ for $x=1$: $g(1) = (1+e^1)^2+1 = (1+e)^2+1$ This matches the given condition, confirming our expression for $g(x)$ is correct.

Step 4: Calculate the value of $g(0)$ Substitute $x=0$ into the expression for $g(x)$: $g(0) = (1+e^0)^2+1$ Since $e^0 = 1$: $g(0) = (1+1)^2+1$ $g(0) = (2)^2+1$ $g(0) = 4+1$ $g(0) = 5$

The final answer is $\boxed{5}$.

Explanation of the solution: The integral $\int \frac{(1+e^x).e^x}{(1+e^x)^2+1}dx$ is solved by substituting $t = 1+e^x$, which leads to $dt = e^x dx$. The integral transforms to $\int \frac{t}{t^2+1} dt$. This is equal to $\frac{1}{2} \ln(t^2+1) + C$. Substituting back $t = 1+e^x$, we get $\frac{1}{2} \ln((1+e^x)^2+1) + C$. Using logarithm properties, this becomes $\ln(\sqrt{(1+e^x)^2+1}) + C$. Comparing this with the given $ln(\sqrt{g(x)})+C$, we find $g(x) = (1+e^x)^2+1$. Finally, evaluate $g(0) = (1+e^0)^2+1 = (1+1)^2+1 = 2^2+1 = 5$.